Order of element

**PvtBillPilgrim** · November 26th 2006, 05:08 PM

How do you prove that the order of a in Zn (where n is an integer greater than or equal to 1) viewed as a group of order n with respect to addition is
n/(gcd(a,n))?

**ThePerfectHacker** · November 26th 2006, 05:59 PM

Originally Posted by PvtBillPilgrim

How do you prove that the order of a in Zn (where n is an integer greater than or equal to 1) viewed as a group of order n with respect to addition is
n/(gcd(a,n))?

You need to show that:

1) $a^{n/\gcd(a,n)}=0$

2)It is the smallest positive exponent which makes this statement true.

**PvtBillPilgrim** · November 27th 2006, 04:37 AM

Do you mind elaborating on this?

By the way, I'm in a university abstract algebra course. He introduces some group theory at the end. Thanks for the help anyway.

**ThePerfectHacker** · November 27th 2006, 07:18 AM

Originally Posted by PvtBillPilgrim

Do you mind elaborating on this?

By the way, I'm in a university abstract algebra course. He introduces some group theory at the end. Thanks for the help anyway.

Let $G$ be a cyclic group (finite). And let $|G|=n$ . It means the $G$ has a generator that is $<a>=G$ for some $a\in G$ .

Let $b\in G$ thus, $b=a^c$ for $1\leq c\leq n$ .

We need to find the smallest $m$ such as,
$b^m=(a^c)^m=a^{cm}=e$
By the properties of cyclic groups and the fact that $a$ is a generator it is equivalent to saying $n$ divides $cm$ .
Thus, we need the smallest $m$ such that,
$\frac{cm}{n}$ is an integer.
We can write it as, (by dividing by $d=\gcd (c,n)$
$\frac{m(c/d)}{(n/d)}$
But,
$\gcd \left( c/d,n/d \right)=1$
Thus,
$n/d$ divides $m$ .
The smallest such $m$ is of course,
$n/d$

Thus, the order of any element is,
$\frac{n}{\gcd(c,n)}$

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